A brute-force approach assuming that the solution is a whole number, and I am understanding the problem correctly, follows after a few blank lines...
((0 = -/ .*@:(2 2$21 38 55 106 - ])"0) # ]) i.111 4 Checking, -/ .* (2 2$21 38 55 106 - 4) 0 Indeed, 17 is a common factor. Actually, +./ 21 38 55 106 - 4 17 On Tue, Aug 14, 2018 at 2:07 AM, Skip Cave <s...@caveconsulting.com> wrote: > Attempting to solve the following Quora problem: > > What number must be subtracted from 21, 38, 55, and 106 each so that the > remainders (technically differences) are proportional? > > Subtract the integers 0-9 from all four integers in the problem. Then find > the prime factors of each set of four integers. Finally look for columns > that have common factors in all four results: > > ]b=.q: each 21 38 55 106-/a=.i.10 > > ┌────┬───────┬────────┬──────┬──────┬───────┬─────────┬───── > ────┬─────┬─────┐ > > │3 7 │2 2 5 │19 │2 3 3 │17 │2 2 2 2│3 5 │2 7 │13 │2 2 3│ > > ├────┼───────┼────────┼──────┼──────┼───────┼─────────┼───── > ────┼─────┼─────┤ > > │2 19│37 │2 2 3 3 │5 7 │2 17 │3 11 │2 2 2 2 2│31 │2 3 5│29 │ > > ├────┼───────┼────────┼──────┼──────┼───────┼─────────┼───── > ────┼─────┼─────┤ > > │5 11│2 3 3 3│53 │2 2 13│3 17 │2 5 5 │7 7 │2 2 2 2 3│47 │2 23 │ > > ├────┼───────┼────────┼──────┼──────┼───────┼─────────┼───── > ────┼─────┼─────┤ > > │2 53│3 5 7 │2 2 2 13│103 │2 3 17│101 │2 2 5 5 │3 3 11 │2 7 7│97 │ > > └────┴───────┴────────┴──────┴──────┴───────┴─────────┴───── > ────┴─────┴─────┘ > > > Now how can I write a J function that lists the column index in this array > where all four sets of factors have at least one common factor? Spoiler - > In the example, the 5th set (column index 4) has a common factor of 17. > > > Skip > > > Skip Cave > Cave Consulting LLC > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
